Question

If \( \vec{a} \cdot \vec{b} = 0 \) and \( \vec{a} + \vec{b} \) makes an angle of \( 60^\circ \) with \( \vec{a} \), then:

• A. \( |\vec{a}| = 2|\vec{b}| \)
• B. \( 2|\vec{a}| = |\vec{b}| \)
• C. \( |\vec{a}| = \sqrt{3}|\vec{b}| \)
• D. \( |\vec{a}| = |\vec{b}| \)
• E. \( \sqrt{3}|\vec{a}| = |\vec{b}| \)

Hint:

Think of this geometrically as a right-angled triangle where the sides are magnitudes \( |\vec{a}| \) and \( |\vec{b}| \). Then \( \tan(\text{angle with } a) = \text{Opposite}/\text{Adjacent} = |\vec{b}|/|\vec{a}| \).

Answer 1

Answer: (E)

Concept: The condition \( \vec{a} \cdot \vec{b} = 0 \) means the vectors are perpendicular. If \( \phi \) is the angle that the resultant \( \vec{a} + \vec{b} \) makes with \( \vec{a} \), then: \[ \tan \phi = \frac{|\vec{b}| \sin \theta}{|\vec{a}| + |\vec{b}| \cos \theta} \] Since \( \theta = 90^\circ \), this simplifies to \( \tan \phi = \frac{|\vec{b}|}{|\vec{a}|} \).

Step 1: Apply the tangent formula for the angle with the resultant.
Given \( \phi = 60^\circ \) and \( \theta = 90^\circ \): \[ \tan 60^\circ = \frac{|\vec{b}|}{|\vec{a}|} \]

Step 2: Substitute the value of \(\tan 60^\circ\).
We know \( \tan 60^\circ = \sqrt{3} \): \[ \sqrt{3} = \frac{|\vec{b}|}{|\vec{a}|} \] \[ |\vec{b}| = \sqrt{3}|\vec{a}| \]